Q. A straight wire carrying a current $10\ A$ is bent into a semicircular arc of radius $5\ cm$. The magnitude of magnetic field at the centre is
A) $1.5 \times 10^{- 5}T$
B) $3.14 \times 10^{- 5}T$
C) $6.28 \times 10^{- 5}T$
D) $19.6 \times 10^{- 5}T$
Explanation
(c) $B = \frac{\mu_{0}}{4\pi} \times \frac{\pi i}{r} \Rightarrow B = 10^{- 7} \times \frac{\pi \times 10}{5 \times 10^{- 2}} = 6.28 \times 10^{- 5}T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.